Curves on P1 × P1
Peter Bruin
16 November 2005
1. Introduction
One of the exercises in last semester’s Algebraic Geometry course went as follows:
Exercise. Let k be a field and Z = P1k ×k P1k . Show that the Picard group Pic Z is the free Abelian
group generated by the classes of a horizontal and a vertical line.
Here Pic Z is to be interpreted as the divisor class group Cl Z, to which it is naturally isomorphic
for Noetherian integral separated locally factorial schemes [Hartshorne, Corollary 6.16]. We view the
first P1 as the result of glueing Spec(k[x]) and Spec(k[1/x]) via Spec(k[x, 1/x]), and similarly for the
second P1 with y instead of x. Then Z = P1k ×k P1k is the result of glueing the spectra of k[x, y],
k[x, 1/y], k[1/x, y] and k[1/x, 1/y] in the obvious way.
To prove the claim (see [Hartshorne, Example II.6.6.1] for a different approach), let Lx and Ly be
the vertical and horizontal lines x = ∞ and y = ∞. More precisely, Lx is determined by the coherent
sheaf of ideals ILx with
Ã
for A = k[x, y] and A = k[x, 1/y]
ILx |Spec A =
1/x · Ã for A = k[1/x, y] and A = k[1/x, 1/y],
and similarly for Ly . If Y is a curve on Z different from Lx and Ly (curves are assumed to be integral),
the intersection of Y with Spec(k[x, y]) is a plane curve defined by an irreducible polynomial f ∈ k[x, y].
Let a be the degree of f as a polynomial in x and b is its degree as a polynomial in y; then the divisor
of f as a rational function on Z equals
(f ) = Y − a · Lx − b · Ly ,
so we see that the divisor class of Y is equal to
[Y ] = a[Lx ] + b[Ly ].
This shows that Cl Z is generated by [Lx ] and [Ly ]; because there are no rational functions f ∈ k(x, y)
with the property that (f ) = a · Lx + b · Ly as a divisor on Z unless a = b = 0, the classes [Lx ]
and [Ly ] are linearly independent. If Y is a divisor on Z and a, b are the unique integers with
[Y ] = a[Lx ] + b[Ly ], we say that Y is of type (a, b).
The isomorphism Cl Z → Pic Z sends the class of a divisor Y of type (a, b) to the invertible sheaf
OZ (Y ) ∼
= OZ (a · Lx + b · Ly ). Note that OZ (a · Lx ) is isomorphic to the pullback p∗1 OP1k (a · ∞) ,
where the invertible sheaf OP1k (a · ∞) on P1k is defined by
OP1k (a · ∞)|Spec k[x] = (k[x])∼
OP1k (a · ∞)|Spec k[1/x] = xa · (k[1/x])∼ .
On the other hand, there is the invertible sheaf OP1k (a) with
OP1k (a)|Spec k[x/y] = y a · (k[x/y])∼
OP1k (a)|Spec k[y/x] = xa · (k[y/x])∼ ,
which is clearly isomorphic to OP1k (a · ∞), so
OZ (a · Lx ) ∼
= p∗1 (OP1k (a)).
Something similar is true for the second projection. Using
OZ (a · Lx + b · Ly) ∼
= OZ (a · Lx ) ⊗OZ (b · Ly )
1
we conclude that OZ (Y ) is isomorphic to the invertible sheaf O(a, b) on Z defined by
O(a, b) = p∗1 OP1k (a) ⊗OZ p∗2 OP1k (b) .
The aim of this talk is to study the cohomology of the sheaves O(a, b) and to derive some
consequences for the kind of curves that exist on Z. We will do the following:
1. Prove the Künneth formula: if X and Y are Noetherian separated schemes over a field k, there
is a natural isomorphism
∼ H(X, F ) ⊗k H(Y, G)
p∗ G) =
H(X ×k Y, p∗ F ⊗O
1
X×k Y
2
for all quasi-coherent sheaves F on X and G on Y .
2. Deduce a connectedness result for closed subschemes and a genus formula for curves on Z.
3. Prove Bertini’s theorem: if X is a non-singular subvariety of Pnk with k an algebraically closed
field, there exists a hyperplane H ⊂ Pnk not containing X such that H ∩ X is a regular scheme.
4. Deduce that if k is algebraically closed field, there exist non-singular curves of type (a, b) on Z
for all a, b > 0.
2. Tensor products of complexes
Let A be a ring, (C, d) a complex of right A-modules and (C ′ , d′ ) a complex of left A-modules, i.e. C
and C ′ are graded A-modules
M
M
C=
C n and C ′ =
C ′n
n∈Z
n∈Z
and d, d are A-module endomorphisms such that dd = 0 and d(C n ) ⊆ C n+1 (similarly for d′ ). Let
C ⊗A C ′ be the usual tensor product, graded in such a way that
M
(C ⊗A C ′ )n =
C p ⊗A C ′q .
′
p+q=n
′
There is a group endomorphism D of C ⊗A C defined by
D(x ⊗ y) = dx ⊗ y + (−1)p x ⊗ d′ y
for x ∈ C p ;
it fulfills D((C ⊗A C ′ )n ) ⊂ (C ⊗A C ′ )n+1 and DD = 0, so ((C ⊗A C ′ ), D) is a complex of Abelian
groups.
For any complex (C, d) of Abelian groups, we write Z(C) for the subgroup of cocycles, B(C) for
the subgroup of coboundaries and H(C) for the cohomology of C:
Z(C) = ker d,
B(C) = im d,
H(C) = Z(C)/B(C).
′
If x and y are cocycles in C and C , respectively, then x ⊗ y is a cocycle in C ⊗A C ′ , because
D(x ⊗ y) = dx ⊗ y + (−1)p x ⊗ d′ y = 0 for x ∈ C p .
This means that there is a natural A-bilinear map
Z(C) × Z(C ′ ) → Z(C ⊗A C ′ )
(x, y) 7→ x ⊗ y.
If either x ∈ B(C) or y ∈ B(C ′ ), then the image of (x, y) under this map is in B(C ⊗A C ′ ), because
for example
dx ⊗ y = D(x ⊗ y) for all x ∈ C, y ∈ Z(C ′ )
This means that we can divide out by the coboundaries in each of the groups and get a natural
A-bilinear map
H(C) × H(C ′ ) → H(C ⊗A C ′ )
and therefore (by the universal property of the tensor product) a natural group homomorphism
γC,C ′ : H(C) ⊗A H(C ′ ) → H(C ⊗A C ′ ).
In the next section we will need the following result:
2
Lemma. Let A be a ring, (C, d) a complex of right A-modules and (C ′ , d′ ) a complex of left Amodules. Assume d = 0. Then H(C) ∼
= C and γC,C ′ induces a natural group homomorphism
C ⊗A H(C ′ ) −→ H(C ⊗A C ′ )
x ⊗ ȳ 7−→ x ⊗ y.
(1)
If C is flat over A, then this map is an isomorphism.
Proof . We only need to prove the last claim. Because C is flat, ker(D) = ker(1 ⊗ d′ ) = C ⊗A ker(d′ ),
so the natural map C ⊗A Z(C ′ ) → Z(C ⊗A C ′ ) is an isomorphism. Furthermore, the image of
C ⊗A B(C ′ ) in C ⊗A Z(C ′ ) corresponds to the subgroup B(C ⊗A C ′ ) under this isomorphism, since
both are generated by elements of the form x ⊗ d′ y with x ∈ C and y ∈ C ′ . This implies the map
defined above is an isomorphism.
3. The Künneth formula
From now on we restrict our attention to the case where A is a field k. Then all complexes have the
structure of k-vector spaces, and all modules are flat. For a treatment without this restriction, see
[Bourbaki]. We will prove the following theorem (note that the previous lemma is a special case of
this):
Theorem (Künneth formula). Let (C, d) and (C ′ , d′ ) be complexes over k. Then the natural
k-linear map
γC,C ′ : H(C) ⊗k H(C ′ ) → H(C ⊗k C ′ )
is an isomorphism.
Proof . Write Z = Z(C), B = B(C), H = H(C) and H ′ = H(C ′ ). Consider the short exact sequence
of complexes defining Z(C) and B(C):
0
/Z
j
d
/C
/ B(1)
/ 0.
Here B(1) denotes the complex B shifted one place to the left, i.e. B(1)n = B n+1 . Taking the tensor
product with C ′ gives a short exact sequence of complexes
0
/ Z ⊗k C ′
j⊗1
/ C ⊗k C ′
d⊗1
/ (B ⊗k C ′ )(1)
/ 0.
We take the cohomology sequence of this short exact sequence. The coboundary map will go from
H(B ⊗k C ′ ) to H(Z ⊗k C ′ ). To find out what it does, we write down the following diagram with exact
rows:
/0
/ (Z ⊗k C ′ )n−1 j⊗1 / (C ⊗k C ′ )n−1 d⊗1 / (B ⊗k C ′ )n
0
D
D
/ (Z ⊗k C ′ )n
0
j⊗1
D
/ (C ⊗k C ′ )n
d⊗1
/ (B ⊗k C ′ )n+1
/ 0.
Because d = 0 on B and because B is flat over k, the kernel of D: (B ⊗k C ′ )n → (B ⊗k C ′ )n+1 equals
ker(D) = ker(1 ⊗ d′ ) ∼
= B ⊗k ker(d′ ),
so ker D is generated by elements of the form dx ⊗ y with x ⊗ y ∈ (C ⊗k C ′ )n−1 such that y ∈ Z(C ′ ).
The image of x ⊗ y ∈ (C ⊗k C ′ )n−1 in (C ⊗k C ′ )n is now D(x ⊗ y) = dx ⊗ y, which is in (Z ⊗k C ′ )n .
We see therefore that the coboundary map sends the class of dx ⊗ y to that of (i ⊗ 1)(dx ⊗ y), where
i: B → Z is the inclusion. In other words, the coboundary map equals H(i ⊗ 1). The long exact
sequence is now
H n (B ⊗k C ′ )
/ H n (Z ⊗k C ′ ) H(j⊗1)/ H n (C ⊗k C ′ )
H(i⊗1)
3
H(d⊗1)
/ H n+1 (B ⊗k C ′ ) H(i⊗1)/ H n+1 (Z ⊗k C ′ ).
We can also take the tensor product with H ′ of the short exact sequence defining H to obtain an
exact sequence
/ B ⊗k H ′ i⊗1 / Z ⊗k H ′ p⊗1 / H ⊗k H ′
/ 0.
0
We connect this sequence with the long exact sequence above via the natural maps
γB,C ′ : B ⊗k H ′ → H(C ⊗k C ′ )
γZ,C ′ : Z ⊗k H ′ → H(C ⊗k C ′ )
γC,C ′ : H ⊗k H ′ → H(C ⊗k C ′ ),
the first two of which are the isomorphisms occurring in the lemma from Section 2. This gives a
commutative diagram with exact rows
(B ⊗k H ′ )n
i⊗1
/ (Z ⊗k H ′ )n
γB,C ′
p⊗1
/ (H ⊗k H ′ )n
γZ,C ′
/0
γC,C ′
H(i⊗1)
/ H n (Z ⊗k C ′ ) H(j⊗1)/ H n (C ⊗k C ′ )
H n (B ⊗k C ′ )
H(d⊗1)
/ H n+1 (B ⊗k C ′ ) H(i⊗1)/ H n+1 (Z ⊗k C ′ )
O
O
γZ,C ′
γB,C ′
/ (B ⊗k H ′ )n+1
0
i⊗1
/ (Z ⊗k H ′ )n+1
The lower right part shows that H(i ⊗ 1) is injective, so H(d ⊗ 1) = 0 by exactness. From the rest of
the diagram we now see that γC,C ′ is an isomorphism.
4. The cohomology of sheaves of the form F ⊗k G
Let X and Y be two compact separated schemes over a field k. Consider the scheme Z = X ×k Y
together with its projection morphisms p1 : Z → X and p2 : Z → Y . Let F and G be quasi-coherent
sheaves on X and Y , respectively. Recall that the pullbacks p∗1 F and p∗2 G of F and G to Z are defined
by
p∗1 F = OZ ⊗p−1 OX p1−1 F
1
p∗2 G
= OZ ⊗p−1 OY p−1
2 G.
2
It is a general fact that the pullback of a quasi-coherent sheaf is quasi-coherent. We use this for p∗1 F
and p∗2 G. Suppose U = Spec A and V = Spec B are affine opens of X and Y , respectively, M is an
A-module such that F |U ∼
= N ∼ . Then the restrictions of
= M ∼ and N is a B-module such that F |V ∼
∗
∗
p1 F and p2 G to the affine open subscheme W = U ×k V = Spec(A ⊗k B) of Z are
p∗ G|W ∼
p∗ F |W ∼
= (p∗ G(W ))∼
= (p∗ F (W ))∼
2
1
1
2
∼
= ((A ⊗k B) ⊗B G(V ))∼
∼ (A ⊗k N )∼ .
=
∼
= ((A ⊗k B) ⊗A F (U ))∼
∼ (B ⊗k M )∼ ,
=
From this we get the following expression for the sheaf p∗1 F ⊗OZ p∗2 G:
∼ ((B ⊗k M ) ⊗A⊗ B (A ⊗k N ))∼
p∗1 F ⊗OZ p∗2 G|W =
k
∼
= (M ⊗k N )∼ .
In particular, we see that
p∗1 F ⊗OZ p∗2 G(U ×k V ) ∼
= F (U ) ⊗k G(V )
for all open affine subschemes U of X and V of Y . It seems therefore useful to introduce the abbreviated
notation
F ⊗k G = p∗1 F ⊗OZ p∗2 G
for quasi-coherent sheaves F on X and G on Y . (To prevent confusion, this notation should only be
used if the sheaves are quasi-coherent.)
We are now going to compare the cohomology of the sheaf F ⊗k G on Z to the cohomology of F
on X and G on Y . This we will do using a variant of Čech cohomology with respect to finite affine
coverings of X, Y and Z.
4
Definition. The unordered Čech complex of a sheaf F of Abelian groups on a topological space X
with respect to an open covering U = {Ui }i∈I is the complex defined by
Y
C n (U, F ) =
F (Ui0 ,...,in )
i0 ,...,in ∈I
where, as usual,
Ui0 ,...,in = Ui0 ∩ . . . ∩ Uin .
The maps d: C
complex:
n
→ C
n+1
are defined using the same formula as for the usual (alternating) Čech



X
n+1
.
(−1)j si0 ,...,ı̂j ,...,in+1 |Ui0 ,...,in+1
d {si0 ,...,in }i0 ,...,in ∈I =


j=0
i0 ,...,in+1 ∈I
Notice that, in contrast to the alternating Čech cohomology, all the C n (U, F ) are non-zero (unless
X = ∅), but that the product occuring in the definition of C n (U, F ) is finite if I is finite.
Let U = {Ui }i∈I and V = {Vj }j∈J be finite coverings by open affine subschemes of X and Y ,
respectively. Because X and Y are separated over k, the intersection of any positive number of such
affines is again affine [Hartshorne, Exercise II.4.3]. We look at the unordered Čech complex of the
sheaf F ⊗k G on Z with respect to the affine open covering U × V. By the property (1) of F ⊗k G and
because I and J are finite,
Y
F ⊗k G Ui0 ×k Vj0 ∩ . . . ∩ Uin ×k Vjn
C n (U ×k V, F ⊗k G) =
(i0 ,j0 ),...,(in ,jn )∈I×J
∼
=
M
M
F (Ui0 ,...,in ) ⊗k G(Vj0 ,...,jn ).
i0 ,...,in ∈I j0 ,...,jn ∈J
Since the tensor product is distributive over direct sums, we see that



M
M
C n (U ×k V, F ⊗k G) ∼
F (Ui0 ,...,in ) ⊗k 
=
i0 ,...,in ∈I
j0 ,...,jn ∈J
∼
= C n (U, F ) ⊗k C n (V, G).
We take the direct sum over all n and conclude that
∞
M
C n (U, F ) ⊗k C n (V, G).
C(U ×k V, F ⊗k G) ∼
=

G(Vj0 ,...,jn )
(2)
n=0
Fact. There exists a natural homotopy equivalence of complexes
∞
M
C n (U, F ) ⊗k C n (V, G) ∼ C(U, F ) ⊗k C(V, G).
n=0
After applying this fact, which follows from the Eilenberg–Zilber theorem [Godement, Théorème
3.9.1], to the right-hand side of (2) and taking cohomology, we obtain a natural isomorphism
∼ H(C(U, F ) ⊗k C(V, G)).
H(C(U ×k V, F ⊗k G)) =
Now the Künneth formula implies that
Ȟ(U ×k V, F ⊗k G) ∼
= Ȟ(U, F ) ⊗k Ȟ(V, G).
If X and Y are Noetherian, then from the fact that the Čech cohomology is isomorphic to the derived
functor cohomology for open affine coverings (the proof of [Hartshorne, Theorem III.4.5] also works
for the unordered Čech cohomology) we get the following theorem:
Theorem. Let X and Y be Noetherian separated schemes over a field k. For all quasi-coherent
sheaves F on X and G on Y , there is a natural isomorphism of k-vector spaces
H(X, F ) ⊗k H(Y, G) ∼
= H(X ×k Y, F ⊗k G).
5
5. Application to the sheaves O(a, b) and curves on P1k ×k P1k
We have seen in Dirard’s talk (see also [Hartshorne, Section III.5]) that for any ring A the cohomology
of the sheaves OX (n) on X = PrA is given by
H 0 (X, OX (n)) ∼
= Sn
H i (X, OX (n)) = 0 for 0 < i < r
H r (X, OX (n)) ∼
= HomA (S−n−r−1 , A)
for all n ∈ Z, where Sn is the component of degree n in S = A[x0 , . . . , xr ]. In particular, for A equal
to the field k and for r = 1,
H 0 (P1k , OP1k (n)) ∼
= k[x0 , x1 ]n
1
1
H (P , OP1 (n)) ∼
= k[x0 , x1 ]∨
k
−n−2
k
The dimensions are therefore equal to
dimk H 0 (P1k , OP1k (n)) = max{n + 1, 0}
dimk H 1 (P1k , OP1k (n)) = max{−n − 1, 0}.
It is now a matter of simple calculations and applying the Künneth formula to find the following table
for the cohomology of the sheaves O(a, b) on Z = P1k ×k P1k :
a ≥ −1,
a ≥ −1,
a ≤ −1,
a ≤ −1,
dimk H 0 (Z, O(a, b))
dimk H 1 (Z, O(a, b))
dimk H 2 (Z, O(a, b))
(a + 1)(b + 1)
0
0
0
0
(a + 1)(−b − 1)
(−a − 1)(b + 1)
0
0
0
0
(a + 1)(b + 1)
b ≥ −1
b ≤ −1
b ≥ −1
b ≤ −1
We can now look at a few applications of this. Let Y be a locally principal closed subscheme of Z,
and let i: Y → Z be the inclusion map, which is a closed immersion. Viewing Y as a divisor on Z, we
have an exact sequence of coherent sheaves:
0
/ OZ (−Y )
/ OZ
/ i∗ OY
/ 0.
The corresponding long exact cohomology sequence is
0
GF
@A
GF
@A
/ H 0 (Z, OZ (−Y ))
/ H 0 (Z, OZ )
/ H 0 (Z, i∗ OY )
/ H 1 (Z, OZ (−Y ))
/ H 1 (Z, OZ )
/ H 1 (Z, i∗ OY )
/ H 2 (Z, OZ (−Y ))
/ H 2 (Z, OZ )
/ H 2 (Z, i∗ OY )
BC
ED
BC
ED
/ 0.
Because i is a closed immersion, we know that
H(Z, i∗ OY ) ∼
= H(Y, OY ).
∼ k, H 1 (Z, OZ ) = 0 and H 2 (Z, OZ ) = 0,
Furthermore, the case a = b = 0 gives us that H 0 (Z, OZ ) =
so the long exact sequence breaks down into two exact sequences
0 −→ H 0 (Z, OZ (−Y )) −→ k −→ H 0 (Y, OY ) −→ H 1 (Z, OZ (−Y )) −→ 0
and
0 −→ H 1 (Y, OY ) −→ H 2 (Z, OZ (−Y )) −→ 0.
6
If Y is of type (a, b) with a, b > 0, then OZ (−Y ) ∼
= O(−a, −b); for these sheaves we have by the
bottom row of the table above
H 0 (Z, OZ (−Y )) = 0, H 1 (Z, OZ (−Y )) = 0,
dimk H 2 (Z, OZ (−Y )) = (−a + 1)(−b + 1) = (a − 1)(b − 1).
Therefore,
H 0 (Y, OY ) ∼
=k
and
dimk H 1 (Y, OY ) = (a − 1)(b − 1)
if a, b > 0.
The interpretation of this is that Y is connected, and if Y is a non-singular curve it has genus
(a − 1)(b − 1).
6. Bertini’s theorem
In this section we study intersections of projective varieties with hyperplanes. A hyperplane H ⊂ Pn
is by definition the zero set of a single homogeneous polynomial f ∈ k[x0 , . . . , xn ] of degree 1. Let V
be the subspace of homogeneous elements of degree 1 in k[x0 , . . . , xn ]. Form the projective space
H = (V \ {0})/k ×
= (k[x0 , . . . , xn ]1 \ {0})/k ×
and view it as a projective variety over k; it is isomorphic to Pnk . Because two non-zero sections of
OPn determine the same hyperplane if and only if one is a multiple of the other by an element of k × ,
there is a canonical bijection between H and the set of hyperplanes in Pnk .
Theorem (Bertini). Let X be a non-singular closed subvariety of Pnk , where k is an algebraically
closed field. Then there exists a hyperplane H ⊂ Pnk , not containing X, such that the scheme H ∩ X
is regular. Moreover, the set of all hyperplanes with this property is an open dense subset of H.
Proof . Consider a closed point x of X. There is an i ∈ {0, 2, . . . , n} such that x is not in the hyperplane
defined by xi ; after renaming the coordinates we may assume i = 0. Then f /x0 is a regular function
in a neighbourhood of x for all f ∈ V , so there is a k-linear map
φx : V → OX,x
f 7→ f /x0 ,
where OX,x is the local ring of X at x. If X is contained in the hyperplane H defined by f , then
φx (f ) = 0; conversely, φx (f ) = 0 means that f vanishes on some open neighbourhood of x in X,
hence on all of X since X is irreducible. We conclude that φx (f ) = 0 ⇐⇒ X ⊆ H. Furthermore,
φx (f ) ∈ mx ⇐⇒ x ∈ H.
Assume X 6⊆ H but x ∈ X ∩ H, so that φx (f ) ∈ mx \ {0}. Then f = φx (f )OX,x is a non-zero
ideal of OX,x contained in mx . Now the local ring of H ∩ X at x is OX,x /f, and its maximal ideal is
n = mx /f. The fact that OX,x is an integral domain and f is a non-zero principal ideal implies that
dim(OX,x /f) = dim(OX,x ) − 1.
Furthermore, n2 = (m2x + f)/f and n/n2 ∼
= mx /(m2x + f). In particular,
dimk n/n2 ≤ dimk mx /m2x
with equality if and only if f ⊆ m2 . Recall that dimk mx /m2x ≥ dim OX,x with equality if and only if
OX,x is a regular local ring. Applying this also to OX,x /f we see that OX,x /f is regular if f 6⊆ m2 (in
which case dimk n/n2 = dim OX,x /f), and not regular if f ⊆ m. Hence OX,x /f is a regular local ring if
and only if φx (f ) ∈ mx \ m2x .
Let Bx ⊂ H be the set of hyperplanes that are defined by an element f ∈ V for which φx (f ) ∈ m2x .
In other words, if we put
φ̄x : V → OX,x /m2x
f 7→ f /x0 mod m2x
7
then
Bx = (ker φ̄x \ {0})/k × ⊆ H.
This is a subvariety of H, the interpretation of which is as follows: a hyperplane H is in Bx if and
only if either H ⊇ X or x ∈ H ∩ X and x is a singular point of H ∩ X. Let us take a closer look at
Bx . We put yi = xi /x0 for 1 ≤ i ≤ n, so that Spec k[y1 , . . . , yn ] is an affine open neighbourhood of x.
Let g1 , . . . , gm ∈ k[y1 , . . . , yn ] be local equations for X, and let (a1 , . . . , an ) be the coordinates of the
point x. Then OX,x is isomorphic to Ap , where
A = (k[y1 , . . . , yn ]/(g1 , . . . , gm )),
p = (y1 − a1 , . . . , yn − an ),
and mx corresponds to pAp under this isomorphism. Furthermore, the k-vector space OX,x /m2x has
dimension
dimk (OX,x /m2x ) = dimk (OX,x /mx ) + dimk (mx /m2x ) = 1 + dim X
and is spanned over k by the elements 1, y1 − a1 , . . . , yn − an (easy check). This shows that φ̄x is
surjective, and
dim ker φ̄x = dimk V − dimk (OX,x /m2x )
= (n + 1) − (1 + dim X)
= n − dim X,
from which we conclude that dim Bx = n − dim X − 1.
The polynomials g1 , . . . , gm which locally define X are modulo m2x congruent to the polynomials
ḡi =
n
X
(yj − aj )
j=1
Because φx (f ) is of the form b0 +
Pn
∂gi
(a1 , . . . , an )
∂yj
j=1 bj yj ,
(1 ≤ i ≤ m).
we see that
φx (f ) ∈ m2x ⇐⇒ f /x0 ∈
m
X
kḡi ,
i=1
or, equivalently,
ker φ̄x =
m
X
kx0 ḡi
and Bx =
m
X
i=1
i=1
!
kx0 ḡi \ {0} /k × .
Consider the fibred product X ×k H. Because of the above characterisation of ker φ̄x , there is a
closed subscheme B of X ×k H such that the closed points of B are precisely the points of X ×k H
corresponding to the pairs (x, H) with x a closed point of X and H ∈ Bx .
We have seen that the fibre of B above each point of X has dimension n − dim X − 1, so B itself
has dimension (n − dim X − 1) + dim X = n − 1. Because X is proper over k and proper morphisms
are preserved under base extension, the projection p2 : X ×k H → H is proper too. This implies that
p2 (B) is a closed subset of H of dimension at most n − 1, and from this we conclude that H − p2 (B)
is an open dense subset of H. For each H ∈ H \ p2 (B), the scheme H ∩ X is regular at every point by
the construction of B.
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7. Application to the existence of non-singular curves of type (a, b)
Let k be an algebraically closed field, and let a, b be positive integers. We want to show that there are
non-singular curves of type (a, b) on P1k ×k P1k . First we embed P1k ×k P1k into Pnk , where n = ab+a+b,
using the a-uple, b-uple and Segre embeddings:
P1k ×k P1k −→ Pak ×k Pak −→ Pnk .
Recall that the a-uple embedding is defined by
(x0 : x1 ) 7→ (xa0 : x0a−1 x1 : . . . : xa1 )
and similarly for the b-uple embedding; the Segre embedding is defined by
((s0 : . . . : sa ), (t0 : . . . : tb )) 7→ (. . . : si tj : . . .)
in lexicographic order. Let j denote the composed embedding P1k ×k P1k → Pnk . The image of j is
a non-singular surface X in Pnk that is isomorphic to P1k ×k P1k . We apply Bertini’s theorem to find
a hyperplane H in PN
k such that H ∩ X is a one-dimensional regular closed subscheme of X. This
hyperplane is given by a homogeneous linear polynomial in the coordinates {zi,j : 0 ≤ i ≤ a, 0 ≤ j ≤ b}
of Pnk . Now
zi,j = j(x0a−i xi1 y0b−j y1j ),
so Y = j −1 (H ∩ X), viewed as a divisor on P1k ×k P1k , is of type (a, b). We have seen earlier that this
implies that Y is connected. The local rings of Y are regular local rings, so in particular they are
integral domains [Hartshorne, Remark II.6.11.1A]. This means that there cannot be two irreducible
components of Y intersecting each other; therefore Y is irreducible, and hence a non-singular curve.
References
[Bourbaki] N. Bourbaki, Algèbre, chapitre 10: Algèbre homologique. Masson, Paris, 1980.
[Godement] R. Godement, Topologie algébrique et théorie des faisceaux . Actualités Sci. Ind. 1252.
Hermann, Paris, 1958.
[Hartshorne] R. Hartshorne, Algebraic Geometry. Graduate Texts in Mathematics 52. SpringerVerlag, New York–Heidelberg, 1977.
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